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Use polar coordinates to find the volume of the given solid. Above the cone $z = x^2 + y^2$ and below the sphere $x^2 + y^2 + z^2 = 81$. I've done this problem 5 times now, but can't seem to get the right answer. I found the bounds to be $0 \le \theta \le 2\pi$ and $0 \le r \le \sqrt{\frac{81}{2}}$. I've done the integration and got -$\frac{243\pi}{\sqrt{2}}$ + 459$\pi$

Can someone please tell me where I went wrong?

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First I must say that $z = x^2 + y^2$ is a paraboloid and not a cone, just for being exact.

Now for the problem: You need to calculate the volume between the paraboloid and the sphere in the domain where the paraboloid is inside the sphere. So first let us find this domain:

The sphere and the paraboloid intersect where $z + z^2 = 81$ (since $z = x^2 + y^2$ and $x^2 + y^2 + z^2 = 81$), which is $z_0 = \frac {\sqrt {325} - 1} 2$, where the paraboloid and the sphere create a circle of radius $\sqrt {z_0}$. To get the volume between the sphere and the paraboloid there we need to calculate $\underset {x^2+y^2 \le z_0} \iint (\sqrt {81-x^2-y^2} - (x^2 + y^2)) \,dx \,dy$

We'll do a change of variables to polar coordinates, and will get: $\underset {[0, z_0] \times [0, 2\pi]} \iint (\sqrt {81-r^2} - r^2)r \,dr \,d\theta = \int_{0}^{2\pi}(\int_{0}^{z_0} (\sqrt {81-r^2} - r^2)r \,dr) \,d\theta = 2\pi(\int_{0}^{z_0} r\sqrt {81-r^2} \,dr - \int_{0}^{z_0} r^3 \,dr) = 2\pi (243-\frac 1 3(81-{z_0}^2)^{\frac 3 2} - \frac {{z_0}^4} 4)$

And this is the wanted volume.

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